What is the derivative of sqrt(x^2+y^2-5) ?

The derivative of sqrt(x^2+y^2-5) is (x+y\frac{dy)/(dx)}{sqrt(x^2+y^2-5)}

What is the first derivative of sqrt(x^2+y^2-5) ?

The first derivative of sqrt(x^2+y^2-5) is (x+y\frac{dy)/(dx)}{sqrt(x^2+y^2-5)}

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\frac{d}{d x} (x^{2} + y^{2} - 5)

\frac{x + y \frac{d y}{d x}}{x ^{2} + y ^{2} - 5}

Solution steps

\frac{d}{d x} (x^{2} + y^{2} - 5)

Apply the chain rule:

\frac{1}{2 x ^{2} + y ^{2} - 5} \frac{d}{d x} (x^{2} + y^{2} - 5)

= \frac{1}{2 x ^{2} + y ^{2} - 5} \frac{d}{d x} (x^{2} + y^{2} - 5)

\frac{d}{d x} (x^{2} + y^{2} - 5) = 2 x + 2 y \frac{d y}{d x}

= \frac{1}{2 x ^{2} + y ^{2} - 5} (2 x + 2 y \frac{d y}{d x})

Simplify

\frac{1}{2 x ^{2} + y ^{2} - 5} (2 x + 2 y \frac{d y}{d x}) : \frac{x + y \frac{d y}{d x}}{x ^{2} + y ^{2} - 5}

= \frac{x + y \frac{d y}{d x}}{x ^{2} + y ^{2} - 5}

\frac{d}{d x} (3 x - \frac{1}{x})

\int_{2}^{6} \frac{x}{x ^{2} + 4 x + 13} d x

f (x) = (2 + 3 x) (6 - 5 x)

x^{2} + 6 x + 8, 0

f (x) = 3 sin (x)

What is the derivative of sqrt(x^2+y^2-5) ?
The derivative of sqrt(x^2+y^2-5) is (x+y\frac{dy)/(dx)}{sqrt(x^2+y^2-5)}
What is the first derivative of sqrt(x^2+y^2-5) ?
The first derivative of sqrt(x^2+y^2-5) is (x+y\frac{dy)/(dx)}{sqrt(x^2+y^2-5)}